 Solution

No. On $\{0,1,2\}$ take $R=\{(0,1),(1,2),(2,0)\}$. This is an antisymmetric relation because it contains no opposite pair of distinct elements. Its reflexive transitive closure contains every ordered pair: following the directed three-cycle reaches any vertex from any other. In particular it contains both $(0,1)$ and $(1,0)$, violating antisymmetry.

This illustrates the reachability preorder: collapsing mutually reachable vertices into strongly connected components yields a partial order, but retaining all original vertices need not.